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The following is a curve in $3$ dimensions:

$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/3) \\ z & = & \cos(\theta - 2\pi/3) \end{eqnarray}$$

Is the curve a circle?

If it is, what about this curve in $4$ dimensions?

$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/4) \\ z & = & \cos(\theta - 2\pi/4) \\ w & = & \cos(\theta - 3\pi/4) \end{eqnarray}$$

I don't know if there is something like a circle in $4$-D. If there is, is this curve the $4$-D version of a circle?

P.S.: Is two-dimensional subspace the generalized plane? I want to learn more about this. What should I read?

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First one: i.imgur.com/tQvqg.png –  user2468 Aug 22 '12 at 17:41
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Probably it is easiest to first show that $x^2+y^2+z^2$ is constant. Then you have show that the points lie on a two-dimensional subspace, which is less obvious. This strategy should work for the 4D case too. –  Rahul Aug 22 '12 at 17:42
    
@JenniferDylan, how did you do this figure? –  Sigur Aug 22 '12 at 17:47
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@Sigur I used k3dsurf.sourceforge.net –  user2468 Aug 22 '12 at 17:49
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@zjk, if you want to take a circle and embed it in a higher-dimensional space without distorting it, you would naturally want it to still lie in a flat two-dimensional portion of the space, in other words, a two-dimensional subspace. Linear algebra would be a useful thing to learn in this context. –  Rahul Aug 22 '12 at 18:56

6 Answers 6

up vote 21 down vote accepted

Here is a simple treatment of the $n$-dimensional case, where the point on the curve is $\vec x = (x_0, \ldots, x_{n-1})$ with $x_i = \cos(\theta-\pi i/n)$.

  1. The curve lies on a sphere.

    We have $x_i^2 = \cos^2(\theta-\pi i/n) = \frac12+\frac12\cos(2\theta-2\pi i/n)$. So $$\|\vec x\|^2 = \sum x_i^2 = \frac n2 + \frac12 \sum \cos(2\theta-2\pi i/n).$$ The latter term is zero because it is the sum of $n$ equally spaced sinusoids (it is equivalently the $x$-component of the sum of $n$ unit vectors equally spaced along the unit circle, or the real part of the sum of the $n$th roots of $e^{2n\theta\sqrt{-1}}$; in either case, the entire thing is zero by symmetry). So $\|\vec x\|^2$ is a constant, $\frac n2$, independent of $\theta$.

  2. The curve lies on a two-dimensional subspace.

    We have $x_i = \cos(\theta-\pi i/n) = a_i\cos\theta + b_i\sin\theta$ for some fixed $a_i$ and $b_i$ independent of $\theta$. Then $\vec x = \vec a\cos\theta + \vec b\sin\theta$. So $\vec x$ lies on the two-dimensional subspace spanned by $\vec a$ and $\vec b$.

Thus, $\vec x$ lies on the intersection of a sphere in $n$ dimensions and a two-dimensional subspace, i.e. a sphere in two dimensions, also known as a circle.

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Using trigonometric identities, we get $$\cos(\theta-\frac\pi3) = \cos\theta\cos\frac\pi3+\sin\theta\sin\frac\pi3=\frac12\cos\theta+\frac12\sqrt3\sin\theta,$$ $$\cos(\theta-\frac{2\pi}3) = \cos\theta\cos\frac{2\pi}3+\sin\theta\sin\frac{2\pi}3=-\frac12\cos\theta+\frac12\sqrt3\sin\theta.$$ Thus, if we write $\gamma=(x,y,z)$ we get $\gamma = v_1\cos\theta + v_2\sin\theta$ with $v_1=(1,\frac12,-\frac12)$ and $v2=(0,\frac12\sqrt3,\frac12\sqrt3)$. Thus, we have at least an ellipse. Moreover, it is easy to check that $v1\cdot v2=0$. Furthermore, $v_1^2 = 1 + \frac14+\frac14=\frac32$ and $v_2^2=\frac34+\frac34=\frac32$; thus the vectors are orthogonal and of equal length, and therefore it is indeed a circle.

In the general case, we have $(v_1)_k = \cos \frac{(k-1)\pi}d$ and $(v_2)_k=\sin \frac{(k-1)\pi}d$, where $d$ is the dimension of the vector space ($d=4$ in your second case). Thus, $$v_1\cdot v_2 = \sum_{k=1}^d\cos\frac{(k-1)\pi}d\sin\frac{(k-1)\pi}d = \frac12\sum_{k=1}^d\sin\frac{(k+1)2\pi}{d} = 0$$ and $$v_2^2-v_1^2 = \sum_{k=1}^d\left(\cos^2\frac{(k+1)\pi}d-\sin^2\frac{(k+1)\pi}d\right) = \sum_{k=1}^d\cos\frac{(k+1)2\pi}{d} = 0,$$ The construction thus gives a circle in any dimension.

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$\cos3\theta=cos3(\theta)$

$\cos3(\theta-\frac{\pi}{3})=cos(3\theta-\pi)=-\cos3\theta$

As $-y=-\cos(\theta-\frac{\pi}{3})=cos(\theta+\frac{2\pi}{3})$,

$\cos3(\theta+\frac{\pi}{3})=\cos(2\pi+3\theta)=\cos3\theta$

$\cos3(\theta-\frac{2\pi}{3})=cos(3\theta-2\pi)=\cos3\theta$

Now, $\cos3\theta=4cos^3\theta-3\cos\theta$

If $\cos3\theta=a$ and $\cos\theta=t$

So, $x,-y,z$ are the roots of $4t^3-3t-a=0$

$=>x+(-y)+z=0$

$=>x(-y)+(-y)z+zx=\frac{3}{4}$

$=>x^2+y^2+z^2=(x+(-y)+z)^2-2(x(-y)+(-y)z+zx)=0+2\frac{3}{4}$

$=>x^2+y^2+z^2=\frac{3}{2}$

Observe that $(x,y,z)$ satisfy a general plane equation $Ax+By+CZ+D=0$ where $A,B,C,D $ constants, not all zeros.

Also, satisfies the equation of the general circle in 3-D, $(x-a)^2+(y-b)^2+(z-c)^2=d^2 $. $a=b=c=0, d^2=\frac{3}{2}$

In case of $x,y,z,w$,

$z=\cos(\theta-\frac{2\pi}{4})=\sin\theta$, $\sqrt2 y=\cos\theta+\sin\theta$, $\sqrt2 w=-\cos\theta+\sin\theta$

So,$x^2+z^2=1$ and $w^2+y^2=1$

$x^2+y^2+z^2+w^2=2$

Now $\sqrt2 (y+w)=2\sin\theta=2z=>\sqrt 2z=y+w$

Similarly, $y-w=\sqrt 2x$

Observe that $(x,y,z,w)$ satisfies two general plane equations $Ax+By+CZ+Dw=E$ where $A,B,C,D,E $ constants, not all zeros.

Also, satisfies the equation of the general circle in 4-D, $(x-a)^2+(y-b)^2+(z-c)^2+(w-d)^2=e^2 $ . $a=b=c=d=0, e^2=2$

Again, we know $\cos nx=$Real part of $(\cos x+i\sin x)^n=(\cos x)^n+^nC_2(\cos x)^{n-2}(\sin x)^2+^nC_4(\cos x)^{n-4}(\sin x)^4+...$

Observe there is no term containing $=(\cos x)^{n-1}$

As $\cos n(2x-\frac{2r_i\pi}{n})=\cos(2nx-2r_i\pi)=\cos 2nx=C(say)$

So, $\cos (2x-\frac{2r_i\pi}{n})=R_i$(say), where all $r_i$s are distinct integers with $0 ≤r_i< n$ are the roots of the equation

$2^{n-1}y^n+C_1y^{n-2}+...-C=0$

So, $\sum R_i=0$ as the coefficient of $y^{n-1}$ is 0.

If $x_i=\cos (x-\frac{r_i\pi}{n})=>R_i=2(x_i)^2-1$

So, $\sum (2(x_i)^2-1)=0 =>\sum (x_i)^2=\frac{n}{2}$

This is another way of generalization("The curve lies on a sphere") already achieved by Rahul Narain.

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You need to put some more words and say that you're trying to prove $x^2 + y^2 + z^2 = 3/2$, and that $x, y, z$ satisfy a plane equation (similar wording for the 4D case). –  user2468 Aug 22 '12 at 18:19

For at least the three-dimensional case, here's a more mechanical method (i.e. less enlightening than Rahul's nice answer) to verify if your curve is a circle:

We try to evaluate the curvature $\kappa$ and torsion $\tau$ of the given curve. By the fundamental theorem of space curves, a space curve is uniquely determined (up to rigid motions) by $\kappa(s)$ and $\tau(s)$; if, in addition, $\tau(s)=0$ (i.e. the curve is flat) and $\kappa(s)$ is a constant greater than zero, then we know that the space curve is indeed a circle.

Using formula 26 here for the curvature, we have

$$\begin{align*} \kappa&=\frac{\|\mathbf r^\prime\times \mathbf r^{\prime\prime}\|}{\|\mathbf r^\prime\|^3}\\ &=\frac{\left\|\langle-\sin\,\theta,\cos\left(\theta+\frac{\pi}{6}\right),\cos\left(\theta-\frac{\pi}{6}\right)\rangle\times\langle-\cos\,\theta,-\sin\left(\theta+\frac{\pi}{6}\right),\sin\left(\frac{\pi}{6}-\theta\right)\rangle\right\|}{\left\|\langle-\sin\,\theta,\cos\left(\theta+\frac{\pi}{6}\right),\cos\left(\theta-\frac{\pi}{6}\right)\rangle\right\|^3}\\ &=\sqrt\frac23 \end{align*}$$

Using formula 3 here for the torsion, we have

$$\begin{align*} \tau&=\frac{\mathbf r^\prime\times \mathbf r^{\prime\prime}\cdot\mathbf r^{\prime\prime\prime}}{\kappa^2}\\ &=\frac1{2/3}\begin{vmatrix}-\sin\,\theta&\cos\left(\theta+\frac{\pi}{6}\right)&\cos\left(\theta-\frac{\pi}{6}\right)\\-\cos\,\theta&-\sin\left(\theta+\frac{\pi}{6}\right)&\sin\left(\frac{\pi}{6}-\theta\right)\\\sin\,\theta&-\cos\left(\theta+\frac{\pi}{6}\right)&-\cos\left(\theta-\frac{\pi}{6}\right)\end{vmatrix}\\ &=0 \end{align*}$$

(Note that the determinant is easily seen to be zero, since the third row is a multiple of the first.)


Since $\tau=0$, the curve is flat; in addition, since $\kappa=\sqrt{2/3}$, we find that our space curve is a circle with radius $1/\kappa=\sqrt{3/2}$.

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It is a circle. If you look at the distance from the origin

$$ r = \sqrt{ \cos^2(\theta) + \cos^2(\theta+\frac{\pi}{3}) + \cos^2(\theta+\frac{2\pi}{3}) } $$

which simplifies to $r = \frac{\sqrt{6}}{2} $.

With the 4-dimensional case the distance simplifies to $r=\sqrt{2}$.

I wonder how to prove the general case of

$$ r^2 = \sum_{i=1}^N \left[ \cos^2\left( \theta + \frac{i-1}{N} \pi \right) \right] = \frac{N}{2} $$

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Constant distance from origin implies a sphere not a circle. See this and this. –  user2468 Aug 22 '12 at 18:11
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@JenniferDylan: Yes this is correct, and if the curve was an offset circle then the distance would not be constant either. I guess I needed to show that the curve was planar also. –  ja72 Aug 22 '12 at 19:36

If you use basic trigonometric identities, you can show the first set of three equations is a piece of this plane: $y-z=x$.

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Also, using trig $x^2 + y^2 + z^2 = 3/2.$ So for all $\theta,$ the distance between $(x, y, z)$ and the origin $= \sqrt{3/2}.$ Putting it together sphere $\bigcap$ plane $=$ circle. –  user2468 Aug 22 '12 at 17:41
    
@JenniferDylan Good way to put it :) –  rschwieb Aug 22 '12 at 17:49

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